If $$K_{sp}$$ of $$Ag_2CO_3$$ is $$8 \times 10^{-12}$$, the molar solubility of $$Ag_2CO_3$$ in 0.1 M $$AgNO_3$$ is:

For the sparingly soluble salt $$Ag_2CO_3$$, the dissolution equilibrium in water is

$$Ag_2CO_3(s) \rightleftharpoons 2\,Ag^+(aq) + CO_3^{2-}(aq)$$

The solubility-product constant (Ksp) expression, by definition, is

$$K_{sp}= [Ag^+]^{\,2}\,[CO_3^{2-}]$$

We are told that $$K_{sp}=8\times10^{-12}$$.

Let the molar solubility of $$Ag_2CO_3$$ in the given medium be $$s\;{\rm mol\,L^{-1}}$$. Dissolving one mole of the salt produces two moles of $$Ag^+$$ and one mole of $$CO_3^{2-}$$, so in symbols we have

$$[CO_3^{2-}] = s$$

and

$$[Ag^+] \text{ coming from the salt} = 2s$$

However, the solution already contains $$0.1\;{\rm M}$$ $$AgNO_3$$, which is a strong electrolyte and supplies an initial concentration

$$[Ag^+]_{\text{initial}} = 0.1\;{\rm M}$$

Therefore the total equilibrium concentration of silver ions is

$$[Ag^+]_{\text{total}} = 0.1 + 2s$$

Substituting these concentrations into the $$K_{sp}$$ expression gives

$$K_{sp} = (0.1 + 2s)^{2}\,(s)$$

The unknown $$s$$ is expected to be very small compared with $$0.1$$ M because the common-ion effect suppresses solubility. Hence it is reasonable to approximate

$$0.1 + 2s \approx 0.1$$

(We shall verify this assumption afterwards.) Making this approximation, the equation simplifies to

$$K_{sp} \approx (0.1)^{2}\,s$$

Now we insert the numerical value of $$K_{sp}$$:

$$8 \times 10^{-12} = (0.1)^{2}\,s$$

Since $$(0.1)^{2}=0.01=1\times10^{-2}$$, we have

$$8 \times 10^{-12} = 1\times10^{-2}\,s$$

Solving for $$s$$,

$$s = \dfrac{8 \times 10^{-12}}{1 \times 10^{-2}}$$

$$s = 8 \times 10^{-10}\;{\rm M}$$

We should now verify the earlier approximation. Twice the solubility is

$$2s = 2 \times 8 \times 10^{-10} = 1.6 \times 10^{-9}\;{\rm M}$$

which is indeed much smaller than $$0.1$$ M, confirming that $$0.1 + 2s \approx 0.1$$ was justified.

Thus the molar solubility of $$Ag_2CO_3$$ in $$0.1$$ M $$AgNO_3$$ is

$$s = 8 \times 10^{-10}\;{\rm mol\,L^{-1}}$$

Hence, the correct answer is Option D.